The alloying elements in tool steels (e.g., Cr, V, W, and Mo) combine with the carbon to form very hard and wear-resistant carbide compounds.

11.4 Compute the volume percent of graphite VGr in a 3.5 wt% C cast iron, assuming that all the carbon exists as the graphite phase. Assume densities of 7.9 and 2.3 g/cm3 for ferrite and graphite, respectively.Solution

We are asked to compute the volume percent graphite in a 3.5 wt% C cast iron. It first becomes necessary to compute mass fractions using the lever rule. From the iron-carbon phase diagram (Figure 11.2), the tie-line in the  and graphite phase field extends from essentially 0 wt% C to 100 wt% C. Thus, for a 3.5 wt% C cast iron

Conversion from weight fraction to volume fraction of graphite is possible using Equation 9.6a as
= 0.111 or 11.1 vol%

11.5 On the basis of microstructure, briefly explain why gray iron is brittle and weak in tension.

Solution

Gray iron is weak and brittle in tension because the tips of the graphite flakes act as points of stress concentration.

White iron--2.5 to 4.0 wt% C and less than 1.0 wt% Si. No heat treatment; however, cooling is rapid during solidification.

Nodular cast iron--2.5 to 4.0 wt% C, 1.0 to 3.0 wt% Si, and a small amount of Mg or Ce. A heat treatment at about 700C may be necessary to produce a ferritic matrix.

(b) With regard to microstructure:

White iron--There are regions of cementite interspersed within pearlite.

Nodular cast iron--Nodules of graphite are embedded in a ferrite or pearlite matrix.

(c) With respect to mechanical characteristics:

White iron--Extremely hard and brittle.

Nodular cast iron--Moderate strength and ductility.

11.8 Is it possible to produce malleable cast iron in pieces having large cross-sectional dimensions? Why or why not?

Solution

It is not possible to produce malleable iron in pieces having large cross-sectional dimensions. White cast iron is the precursor of malleable iron, and a rapid cooling rate is necessary for the formation of white iron, which may not be accomplished at interior regions of thick cross-sections.

Nonferrous Alloys11.9 What is the principal difference between wrought and cast alloys?Solution

The principal difference between wrought and cast alloys is as follows: wrought alloys are ductile enough so as to be hot or cold worked during fabrication, whereas cast alloys are brittle to the degree that shaping by deformation is not possible and they must be fabricated by casting.

11.10 Why must rivets of a 2017 aluminum alloy be refrigerated before they are used?Solution

Rivets of a 2017 aluminum alloy must be refrigerated before they are used because, after being solution heat treated, they precipitation harden at room temperature. Once precipitation hardened, they are too strong and brittle to be driven.

11.11 What is the chief difference between heat-treatable and non-heat-treatable alloys?Solution

The chief difference between heat-treatable and nonheat-treatable alloys is that heat-treatable alloys may be strengthened by a heat treatment wherein a precipitate phase is formed (precipitation hardening) or a martensitic transformation occurs. Nonheat-treatable alloys are not amenable to strengthening by such treatments.

For sand casting, sand is the mold material, a two-piece mold is used, ordinarily the surface finish is not an important consideration, the sand may be reused (but the mold may not), casting rates are low, and large pieces are usually cast.

For die casting, a permanent mold is used, casting rates are high, the molten metal is forced into the mold under pressure, a two-piece mold is used, and small pieces are normally cast.

For investment casting, a single-piece mold is used, which is not reusable; it results in high dimensional accuracy, good reproduction of detail, and a fine surface finish; and casting rates are low.

For lost foam casting, the pattern is polystyrene foam, whereas the mold material is sand. Complex geometries and tight tolerances are possible. Casting rates are higher than for investment, and there are few environmental wastes.

For continuous casting, at the conclusion of the extraction process, the molten metal is cast into a continuous strand having either a rectangular or circular cross-section; these shapes are desirable for subsequent secondary metal-forming operations. The chemical composition and mechanical properties are relatively uniform throughout the cross-section.

Miscellaneous Techniques11.17 If it is assumed that, for steel alloys, the average cooling rate of the heat-affected zone in the vicinity of a weld is 10°C/s, compare the microstructures and associated properties that will result for 1080 (eutectoid) and 4340 alloys in their HAZs.Solution

This problem asks that we specify and compare the microstructures and mechanical properties in the heat-affected weld zones for 1080 and 4340 alloys assuming that the average cooling rate is 10C/s. Figure 10.27 shows the continuous cooling transformation diagram for an iron-carbon alloy of eutectoid composition (1080), and, in addition, cooling curves that delineate changes in microstructure. For a cooling rate of 10C/s (which is less than 35C/s) the resulting microstructure will be totally pearlite--probably a reasonably fine pearlite. On the other hand, in Figure 10.28 is shown the CCT diagram for a 4340 steel. From this diagram it may be noted that a cooling rate of 10C/s produces a totally martensitic structure. Pearlite is softer and more ductile than martensite, and, therefore, is most likely more desirable.

11.18 Describe one problem that might exist with a steel weld that was cooled very rapidly.Solution

If a steel weld is cooled very rapidly, martensite may form, which is very brittle. In some situations, cracks may form in the weld region as it cools.

Annealing Processes11.19 In your own words describe the following heat treatment procedures for steels and, for each, the intended final microstructure: full annealing, normalizing, quenching, and tempering.Solution

Full annealing--Heat to about 50C above the A3 line, Figure 11.10 (if the concentration of carbon is less than the eutectoid) or above the A1 line (if the concentration of carbon is greater than the eutectoid) until the alloy comes to equilibrium; then furnace cool to room temperature. The final microstructure is coarse pearlite.

Normalizing--Heat to at least 55C above the A3 line Figure 11.10 (if the concentration of carbon is less than the eutectoid) or above the Acm line (if the concentration of carbon is greater than the eutectoid) until the alloy completely transforms to austenite, then cool in air. The final microstructure is fine pearlite.

Quenching--Heat to a temperature within the austenite phase region and allow the specimen to fully austenitize, then quench to room temperature in oil or water. The final microstructure is martensite.

Tempering--Heat a quenched (martensitic) specimen, to a temperature between 450 and 650C, for the time necessary to achieve the desired hardness. The final microstructure is tempered martensite.

11.20 Cite three sources of internal residual stresses in metal components. What are two possible adverse consequences of these stresses?

Solution

Three sources of residual stresses in metal components are plastic deformation processes, nonuniform cooling of a piece that was cooled from an elevated temperature, and a phase transformation in which parent and product phases have different densities.

Two adverse consequences of these stresses are distortion (or warpage) and fracture.

11.21 Give the approximate minimum temperature at which it is possible to austenitize each of the following iron–carbon alloys during a normalizing heat treatment: (a) 0.20 wt% C, (b) 0.76 wt% C, and (c) 0.95 wt% C.

Solution

(a) For 0.20 wt% C, heat to at least 905C (1660F) since the A3 temperature is 850C (1560F).

(b) For 0.76 wt% C, heat to at least 782C (1440F) since the A3 temperature is 727C (1340F).

(c) For 0.95 wt% C, heat to at least 840C (1545F) since the Acm temperature is 785C (1445F).

11.22 Give the approximate temperature at which it is desirable to heat each of the following iron–carbon alloys during a full anneal heat treatment: (a) 0.25 wt% C, (b) 0.45 wt% C, (c) 0.85 wt% C, and (d) 1.10 wt% C.
Solution

(a) For 0.25 wt% C, heat to about 880C (1510F) since the A3 temperature is 830C (1420F).

(b) For 0.45 wt% C, heat to about 830C (1525F) since the A3 temperature is 780C (1435F).

(c) For 0.85 wt% C, heat to about 777C (1430F) since the A1 temperature is 727C (1340F).

(d) For 1.10 wt% C, heat to about 777C (1430F) since the A1 temperature is 727C (1340F).

11.23 What is the purpose of a spheroidizing heat treatment? On what classes of alloys is it normally used?

Solution

The purpose of a spheroidizing heat treatment is to produce a very soft and ductile steel alloy having a spheroiditic microstructure. It is normally used on medium- and high-carbon steels, which, by virtue of carbon content, are relatively hard and strong.

Heat Treatment of Steels11.24 Briefly explain the difference between hardness and hardenability.Solution

Hardness is a measure of a material's resistance to localized surface deformation, whereas hardenability is a measure of the depth to which a ferrous alloy may be hardened by the formation of martensite. Hardenability is determined from hardness tests.

11.25 What influence does the presence of alloying elements (other than carbon) have on the shape of a hardenability curve? Briefly explain this effect.Solution

The presence of alloying elements (other than carbon) causes a much more gradual decrease in hardness with position from the quenched end for a hardenability curve. The reason for this effect is that alloying elements retard the formation of pearlitic and bainitic structures which are not as hard as martensite.

11.26 How would you expect a decrease in the austenite grain size to affect the hardenability of a steel alloy? Why?Solution

A decrease of austenite grain size will decrease the hardenability. Pearlite normally nucleates at grain boundaries, and the smaller the grain size, the greater the grain boundary area, and, consequently, the easier it is for pearlite to form.

11.27 Name two thermal properties of a liquid medium that will influence its quenching effectiveness.Solution

The two thermal properties of a liquid medium that influence its quenching effectiveness are thermal conductivity and heat capacity.

11.28 Construct radial hardness profiles for the following:

(a) A 50-mm (2-in.) diameter cylindrical specimen of an 8640 steel alloy that has been quenched in moderately agitated oilSolution

In the manner of Example Problem 11.1, the equivalent distances and hardnesses tabulated below were determined from Figures 11.14 and 11.17b.

Radial Equivalent HRC

PositionDistance, mmHardness

Surface 7 54

3/4 R 11 50

Midradius 14 45

Center 16 44
The resulting hardness profile is plotted below.
(b) A 75-mm (3-in.) diameter cylindrical specimen of a 5140 steel alloy that has been quenched in moderately agitated oilSolution

In the manner of Example Problem 11.1, the equivalent distances and hardnesses tabulated below were determined from Figures 11.14 and 11.17b.

Radial Equivalent HRC

PositionDistance, mmHardness

Surface 13 41

3/4 R 17.5 37

Midradius 22 33

Center 25 32
The resulting hardness profile is plotted below.
(c) A 65-mm (2-in.) diameter cylindrical specimen of an 8620 steel alloy that has been quenched in moderately agitated waterSolution

In the manner of Example Problem 11.1, the equivalent distances and hardnesses tabulated below were determined from Figures 11.15 and 11.17a.

Radial Equivalent HRC

PositionDistance, mmHardness

Surface 2.5 42

3/4 R 7 31

Midradius 11 25

Center 13 24
The resulting hardness profile is plotted below.

(d) A 70-mm (2-in.) diameter cylindrical specimen of a 1040 steel alloy that has been quenched in moderately agitated water.Solution

In the manner of Example Problem 11.1, the equivalent distances and hardnesses tabulated below were determined from Figures 11.14 and 11.17a.

Radial Equivalent HRC

PositionDistance, mmHardness

Surface 3 48

3/4 R 8 30

Midradius 13 23

Center 15 22
The resulting hardness profile is plotted below.

11.29 Compare the effectiveness of quenching in moderately agitated water and oil by graphing, on a single plot, radial hardness profiles for 65-mm (2-in.) diameter cylindrical specimens of an 8630 steel that have been quenched in both media.

Solution

We are asked to compare the effectiveness of quenching in moderately agitated water and oil by graphing, on a single plot, hardness profiles for a 65 mm (2-1/2 in.) diameter cylindrical specimen of an 8630 steel that has been quenched in both media.

For moderately agitated water, the equivalent distances and hardnesses for the several radial positions [Figures 11.17a and 11.15] are tabulated below.
Radial Equivalent HRC

PositionDistance, mmHardness

Surface 2.5 52

3/4 R 7 43

Midradius 11 36

Center 13 33
While for moderately agitated oil, the equivalent distances and hardnesses for the several radial positions [Figures 11.17b and 11.15] are tabulated below.
Radial Equivalent HRC

PositionDistance, mmHardness

Surface 10 37

3/4 R 15 32

Midradius 18 29

Center 20 28
These data are plotted here.

Precipitation Hardening11.30 Compare precipitation hardening (Section 11.9) and the hardening of steel by quenching and tempering (Sections 10.5, 10.6, and 10.8) with regard to

(a) The total heat treatment procedure

(b) The microstructures that develop

(c) How the mechanical properties change during the several heat treatment stagesSolution

(a) With regard to the total heat treatment procedure, the steps for the hardening of steel are as follows:

(g) High-temperature furnace elements to be used in oxidizing atmospheresSolution

(a) Gray cast iron would be the best choice for an engine block because it is relatively easy to cast, is wear resistant, has good vibration damping characteristics, and is relatively inexpensive.

(b) Stainless steel would be the best choice for a heat exchanger to condense steam because it is corrosion resistant to the steam and condensate.

(c) Titanium alloys are the best choice for high-speed aircraft jet engine turbofan blades because they are light weight, strong, and easily fabricated very resistant to corrosion. However, one drawback is their cost.

(d) A tool steel would be the best choice for a drill bit because it is very hard retains its hardness at high temperature and is wear resistant, and, thus, will retain a sharp cutting edge.

(e) For a cryogenic (low-temperature) container, an aluminum alloy would be the best choice; aluminum alloys have an FCC crystal structure, and therefore, are ductile at very low temperatures.

(f) As a pyrotechnic in flares and fireworks, magnesium is the best choice because it ignites easily and burns readily in air with a very bright flame.

(g) Platinum is the best choice for high-temperature furnace elements to be used in oxidizing atmospheres because it is very ductile, has a relatively very high melting temperature, and is highly resistant to oxidation.

11.D2 A group of new materials are the metallic glasses (or amorphous metals). Write an essay about these materials in which you address the following issues: (1) compositions of some of the common metallic glasses, (2) characteristics of these materials that make them technologically attractive, (3) characteristics that limit their utilization, (4) current and potential uses, and (5) at least one technique that is used to produce metallic glasses.Solution

(a) Compositionally, the metallic glass materials are rather complex; several compositions are as follows: Fe80B20, Fe72Cr8P13C7, Fe67Co18B14Si, Pd77.5Cu6.0Si16.5, and Fe40Ni38Mo4B18.

(b) These materials are exceptionally strong and tough, extremely corrosion resistant, and are easily magnetized.

(c) Principal drawbacks for these materials are 1) complicated and exotic fabrication techniques are required; and 2) inasmuch as very rapid cooling rates are required, at least one dimension of the material must be small--i.e., they are normally produced in ribbon form.

This question provides us with a list of several metal alloys, and then asks us to pick those that may be strengthened by heat treatment, by cold work, or both. Those alloys that may be heat treated are either those noted as "heat treatable" (Tables 11.6 through 11.9), or as martensitic stainless steels (Table 11.4). Alloys that may be strengthened by cold working must not be exceptionally brittle, and, furthermore, must have recrystallization temperatures above room temperature (which immediately eliminates lead). The alloys that fall within the three classifications are as follows:

11.D4 A structural member 100 mm (4 in.) long must be able to support a load of 50,000 N (11,250 lbf) without experiencing any plastic deformation. Given the following data for brass, steel, aluminum, and titanium, rank them from least to greatest weight in accordance with these criteria.

Alloy

Yield Strength

[MPa (ksi)]

Density

(g/cm3)

Brass

415 (60)

8.5

Steel

860 (125)

7.9

Aluminum

310 (45)

2.7

Titanium

550 (80)

4.5

Solution

This problem asks us to rank four alloys (brass, steel, titanium, and aluminum), from least to greatest weight for a structural member to support a 50,000 N (11,250 lbf) load without experiencing plastic deformation. From Equation 6.1, the cross-sectional area (A0) must necessarily carry the load (F) without exceeding the yield strength (y), as

Now, given the length l, the volume of material required (V) is just
Finally, the mass of the member (m) is
Here is the density. Using the values given for these alloys
Thus, titanium would have the minimum weight (or mass), followed by aluminum, steel, and brass.

11.D5 Discuss whether it would be advisable to hot work or cold work the following metals and alloys on the basis of melting temperature, oxidation resistance, yield strength, and degree of brittleness: tin, tungsten, aluminum alloys, magnesium alloys, and a 4140 steel.

Solution

Tin would almost always be hot-worked. Even deformation at room temperature would be considered hot-working inasmuch as its recrystallization temperature is below room temperature (Table 7.2).

Tungsten is hard and strong at room temperature, has a high recrystallization temperature, and experiences oxidation at elevated temperatures. Cold-working is difficult because of its strength, and hot-working is not practical because of oxidation problems. Most tungsten articles are fabricated by powder metallurgy, or by using cold-working followed by annealing cycles.

Most aluminum alloys may be cold-worked since they are ductile and have relatively low yield strengths.

A 4140 steel could be cold-worked in an over-tempered state which leaves it soft and relatively ductile, after which quenching and tempering heat treatments may be employed to strengthen and harden it. This steel would probably have a relatively high recrystallization temperature, and hot-working may cause oxidation.

Heat Treatment of Steels11.D6 A cylindrical piece of steel 25 mm (1.0 in.) in diameter is to be quenched in moderately agitated oil. Surface and center hardnesses must be at least 55 and 50 HRC, respectively. Which of the following alloys will satisfy these requirements: 1040, 5140, 4340, 4140, and 8640? Justify your choice(s).Solution

In moderately agitated oil, the equivalent distances from the quenched end for a one-inch diameter bar for surface and center positions are 3 mm (1/8 in.) and 8 mm (11/32 in.), respectively [Figure 11.17b]. The hardnesses at these two positions for the alloys cited (as determined using Figure 11.14) are given below.

Surface Center

AlloyHardness (HRC)Hardness (HRC)

1040 50 30

5140 56 49

4340 57 57

4140 57 55

8640 57 53

Thus, alloys 4340, 4140, and 8640 will satisfy the criteria for both surface and center hardnesses.

11.D7 A cylindrical piece of steel 75 mm (3 in.) in diameter is to be austenitized and quenched such that a minimum hardness of 40 HRC is to be produced throughout the entire piece. Of the alloys 8660, 8640, 8630, and 8620, which will qualify if the quenching medium is(a) moderately agitated water, and (b) moderately agitated oil? Justify your choice(s).

Solution

(a) This problem calls for us to decide which of 8660, 8640, 8630, and 8620 alloys may be fabricated into a cylindrical piece 75 mm (3 in.) in diameter which, when quenched in mildly agitated water, will produce a minimum hardness of 40 HRC throughout the entire piece.

The center of the steel cylinder will cool the slowest and therefore will be the softest. In moderately agitated water the equivalent distance from the quenched end for a 75 mm diameter bar for the center position is about 17 mm (11/16 in.) [Figure 11.17a]. The hardnesses at this position for the alloys cited (Figure 11.15) are given below.
Center

AlloyHardness (HRC)

8660 58

8640 42

8630 30

8620 22
Therefore, only 8660 and 8640 alloys will have a minimum of 40 HRC at the center, and therefore, throughout the entire cylinder.

(b) This part of the problem asks us to do the same thing for moderately agitated oil. In moderately agitated oil the equivalent distance from the quenched end for a 75 mm diameter bar at the center position is about 25.5 mm (1-1/32 in.) [Figure 11.17b]. The hardnesses at this position for the alloys cited (Figure 11.15) are given below.

Center

AlloyHardness (HRC)

8660 53

8640 37

8630 26

8620 < 20
Therefore, only the 8660 alloy will have a minimum of 40 HRC at the center, and therefore, throughout the entire cylinder.

11.D8 A cylindrical piece of steel 38 mm (1in.) in diameter is to be austenitized and quenched such that a microstructure consisting of at least 80% martensite will be produced throughout the entire piece. Of the alloys 4340, 4140, 8640, 5140, and 1040, which will qualify if the quenching medium is (a) moderately agitated oil and (b) moderately agitated water? Justify your choice(s).

Solution

(a) Since the cooling rate is lowest at the center, we want a minimum of 80% martensite at the center position. From Figure 11.17b, the cooling rate is equal to an equivalent distance from the quenched end of 12 mm (1/2 in.). According to Figure 11.14, the hardness corresponding to 80% martensite for these alloys is 50 HRC. Thus, all we need do is to determine which of the alloys have a 50 HRC hardness at an equivalent distance from the quenched end of 12 mm (1/2 in.). At an equivalent distance of 12 mm (1/2 in.), the following hardnesses are determined from Figure 11.14 for the various alloys.

AlloyHardness (HRC)

4340 56

4140 53

8640 49

5140 43

1040 25
Thus, only alloys 4340 and 4140 will qualify.

(b) For moderately agitated water, the cooling rate at the center of a 38 mm diameter specimen is 7 mm (5/16 in.) equivalent distance from the quenched end [Figure 11.17a]. At this position, the following hardnesses are determined from Figure 11.14 for the several alloys.
AlloyHardness (HRC)

4340 57

4140 55

8640 54

5140 51

1040 33
It is still necessary to have a hardness of 50 HRC or greater at the center; thus, alloys 4340, 4140, 8640, and 5140 qualify.

11.D9 A cylindrical piece of steel 90 mm (3 in.) in diameter is to be quenched in moderately agitated water. Surface and center hardnesses must be at least 55 and 40 HRC, respectively. Which of the following alloys will satisfy these requirements: 1040, 5140, 4340, 4140, 8620, 8630, 8640, and 8660? Justify your choices.Solution

A ninety-millimeter (three and one-half inch) diameter cylindrical steel specimen is to be quenched in moderately agitated water. We are to decide which of eight different steels will have surface and center hardnesses of at least 55 and 40 HRC, respectively.

In moderately agitated water, the equivalent distances from the quenched end for a 90 mm diameter bar for surface and center positions are 3 mm (1/8 in.) and 22 mm (7/8 in.), respectively [Figure 11.17a]. The hardnesses at these two positions for the alloys cited are given below. The hardnesses at these two positions for the alloys cited are given below (as determined from Figures 11.14 and 11.15).
Surface Center

11.D10 A cylindrical piece of 4140 steel is to be austenitized and quenched in moderately agitated oil. If the microstructure is to consist of at least 50% martensite throughout the entire piece, what is the maximum allowable diameter? Justify your answer.

Solution

From Figure 11.14, the equivalent distance from the quenched end of a 4140 steel to give 50% martensite (or a 42.5 HRC hardness) is 27 mm (1-1/8 in.). Thus, the quenching rate at the center of the specimen should correspond to this equivalent distance. Using Figure 11.17b, the center specimen curve takes on a value of 27 mm (1-1/8 in.) equivalent distance at a diameter of about 83 mm (3.3 in.).

11.D11 A cylindrical piece of 8640 steel is to be austenitized and quenched in moderately agitated oil. If the hardness at the surface of the piece must be at least 49 HRC, what is the maximum allowable diameter? Justify your answer.Solution

We are to determine, for a cylindrical piece of 8640 steel, the minimum allowable diameter possible in order yield a surface hardness of 49 HRC, when the quenching is carried out in moderately agitated oil.

From Figure 11.15, the equivalent distance from the quenched end of an 8640 steel to give a hardness of 49 HRC is about 12 mm (15/32 in.). Thus, the quenching rate at the surface of the specimen should correspond to this equivalent distance. Using Figure 11.17b, the surface specimen curve takes on a value of 12 mm equivalent distance at a diameter of about 75 mm (3 in.).

11.D12 Is it possible to temper an oil-quenched 4140 steel cylindrical shaft 100 mm (4 in.) in diameter so as to give a minimum tensile strength of 850 MPa (125,000 psi) and a minimum ductility of 21%EL? If so, specify a tempering temperature. If this is not possible, then explain why.

Solution

This problem asks if it is possible to temper an oil-quenched 4140 steel cylindrical shaft 100 mm (4 in.) in diameter so as to give a minimum tensile strength of 850 MPa (125,000 psi) and a minimum ductility of 21%EL. In order to solve this problem it is necessary to use Figures 11.20a and 11.20c, which plot, respectively, tensile strength and ductility versus tempering temperature. For the 100 mm diameter line of Figure 11.20a, tempering temperatures less than about 560°C are required to give a tensile strength of at least 850 MPa. Furthermore, from Figure 11.20c, for the 100 mm diameter line, tempering temperatures greater than about 585°C will give ductilities greater than 21%EL. Hence, it isnotpossible to temper this alloy to produce the stipulated minimum tensile strength and ductility. To meet the tensile strength minimum, T(tempering) < 560°C, whereas for ductility minimum, T(tempering) > 585°C; thus, there is no overlap of these tempering temperature ranges.

11.D13 Is it possible to temper an oil-quenched 4140 steel cylindrical shaft 12.5 mm (0.5 in.) in diameter so as to give a minimum yield strength of 1000 MPa (145,000 psi) and a minimum ductility of 16%EL? If so, specify a tempering temperature. If this is not possible, then explain why.Solution

This problem asks if it is possible to temper an oil-quenched 4140 steel cylindrical shaft 12.5 mm (0.5 in.) in diameter so as to give a minimum yield strength of 1000 MPa (145,000 psi) and a minimum ductility of 16%EL. In order to solve this problem it is necessary to use Figures 11.20b and 11.20c, which plot, respectively, yield strength and ductility versus tempering temperature. For the 12.5 mm diameter line of Figure 11.20b, tempering temperatures less than about 600°C are required to give a yield strength of at least 1000 MPa. Furthermore, from Figure 11.20c, for the 12.5 mm diameter line, tempering temperatures greater than about 550°C will give ductilities greater than 17%EL. Hence, it ispossible to temper this alloy to produce the stipulated minimum yield strength and ductility; the tempering temperature will lie between 550°C and 600°C.

Precipitation Hardening11.D14 Copper-rich copper–beryllium alloys are precipitation hardenable. After consulting the portion of the phase diagram (Figure 11.30), do the following:

(a) Specify the range of compositions over which these alloys may be precipitation hardened.

(b) Briefly describe the heat-treatment procedures (in terms of temperatures) that would be used to precipitation harden an alloy having a composition of your choosing, yet lying within the range given for part (a).Solution

This problem is concerned with the precipitation-hardening of copper-rich Cu-Be alloys. It is necessary for us to use the Cu-Be phase diagram (Figure 11.30), which is shown below.

(a) The range of compositions over which these alloys may be precipitation hardened is between approximately 0.2 wt% Be (the maximum solubility of Be in Cu at about 300C) and 2.7 wt% Be (the maximum solubility of Be in Cu at 866C).

(b) The heat treatment procedure, of course, will depend on the composition chosen. First of all, the solution heat treatment must be carried out at a temperature within the  phase region, after which, the specimen is quenched to room temperature. Finally, the precipitation heat treatment is conducted at a temperature within the  + 2 phase region.

For example, for a 1.5 wt% Be-98.5 wt% Cu alloy, the solution heat treating temperature must be between about 600C (1110F) and 900C (1650F), while the precipitation heat treatment would be below 600C (1110F), and probably above 300C (570F). Below 300C, diffusion rates are low, and heat treatment times would be relatively long.

11.D15 A solution heat-treated 2014 aluminum alloy is to be precipitation hardened to have a minimum tensile strength of 450 MPa (65,250 psi) and a ductility of at least 15%EL. Specify a practical precipitation heat treatment in terms of temperature and time that would give these mechanical characteristics. Justify your answer.

Solution

We are asked to specify a practical heat treatment for a 2014 aluminum alloy that will produce a minimum tensile strength of 450 MPa (65,250 psi), and a minimum ductility of 15%EL. From Figure 11.27a, the following heat treating temperatures and time ranges are possible to the give the required tensile strength.

Temperature (C)Time Range (h)

260 0.02-0.2

204 0.02-10

149 3-600

121 > 35-?
With regard to temperatures and times to give the desired ductility [Figure 11.27b]:
Temperature (C)Time Range (h)

260 < 0.01, > 40

204 < 0.15

149 < 10

121 < 500
From these tabulations, the following may be concluded:

It is not possible to heat treat this alloy at 260C so as to produce the desired set of properties— there is no overlap of the two sets of time ranges.

At 204C, the heat treating time would be between 0.02 and 0.15 h; times lying within this range are impractically short.

At 149C, the time would be between 3 and 10 h.

Finally, at 121C, the time range is 35 to about 500 h.

11.D16 Is it possible to produce a precipitation-hardened 2014 aluminum alloy having a minimum tensile strength of 425 MPa (61,625 psi) and a ductility of at least 12%EL? If so, specify the precipitation heat treatment. If it is not possible, explain why.

Solution

This problem inquires as to the possibility of producing a precipitation-hardened 2014 aluminum alloy having a minimum tensile strength of 425 MPa (61,625 psi) and a ductility of at least 12%EL. In order to solve this problem it is necessary to consult Figures 11.27a and 11.27b. Below are tabulated the times required at the various temperatures to achieve the stipulated tensile strength.

Temperature (C)Time Range (h)

260 < 0.5

204 < 15

149 1-1000

121 > 35-?
With regard to temperatures and times to give the desired ductility:
Temperature (C)Time Range (h)

260 < 0.02, > 10

204 < 0.4, > 350

149 < 20

121 < 1000
From these tabulations, the following may be concluded:

At 260C, the heat treating time would need to be less than 0.02 h (1.2 min), which is impractically short.

At 204C, the heat treatment would need to be less than 0.4 h (24 min), which is a little on the short side.

At 149C, the time range would be between 1 and 20 h.

Finally, at 121C, this property combination is possible for virtually all times less than about 1000 h.

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